DC Circuit Construction Virtual Lab | Exploring Series and Parallel Circuits

DC Circuit Construction Virtual Lab | Exploring Series and Parallel Circuits

Do have a look at a brief discussion we had around the concept of “Series and Parallel Circuit”

Reflection

  1. Why the brightness of the light source in parallel configuration is more than that in series? How will you go about explaining the phenomena in your words?
  2. On increasing the resistance of one of the three light sources in parallel configuration, the brightness was decreased. How do you explain this? Will similar effect be observed in series configuration too?
  3. In what ways, the model shown in the simulation is different if a similar setup is made in using material components?
  4. Also, do share any thoughts or questions you had while exploring the simulation or during video discussion.

We would also like to get feedback if this was useful and how can be collectively make it better.

Acknowledgements

Discussion
Collaborators in video resource creation (chatshaala): @G_N, @AsmitaRe , @KiranyadavR , @Ashish_Pardeshi , @ravi312

Credits
PhET Interactive Simulations
University of Colorado Boulder
https://phet.colorado.edu

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The simulation platform is really cool and nice. I also liked the interactive session. I just want to reflect upon the answers of the questions in the reflection section,

  1. In series combination, the voltage across the battery is 9 V and both the bulbs have resistance 10 ohms. Now they are in series, so the total resistance of the circuit is (10 + 10) = 20 ohms. Hence from Ohm Law, the current through the circuit will be = (9/20) A = 0.45 A.

Now the power consumption at each of the bulb from Joule’s Law is = [(0.45 * 0.45) * 10] Watts = 2.025 Watts

Similarly, for parallel connection, the voltage across battery and each of the branch is 9 V.
The resistance of each of the branch is = the resistance of the bulb = 10 ohms.
Hence, from Ohm Law the current through each bulb = (9/10) A = 0.9 A

Now the power consumption at each of the bulb from Joule’s Law is = [(0.9 * 0.9) * 10] Watts = 8.1 Watts

As the power consumption at the parallel case is more; hence the brightness of the light source in the parallel source is more than that in series.

  1. If we increase the resistance of one of the bulb in parallel circuit, the current will decrease as the voltage across the bulb is constant. Now suppose we increase the resistance by x times. Then current will also decrease by x times as according to Ohm’s Law IR = V (which is constant here).

From Joule’s Law the power consumption at the bulb will be [(I/x)^2](Rx) = [(I^2)*R]/x ; which is (1/x) times of the former power consumption. That is why, the brightness was decreased.

Yes even in series configuration we increase the resistance of one bulb by x times, then the current will decrease by (1 + x) times (in the case of two bulbs in series).
And hence the power consumption will be [x/{(1 + x)**2}] times of the former case.
And hence brightness will decrease.

  1. In the simulation we used ideal condition; like, zero resistance wire, no internal resistance in battery, no error in the resistances of bulb etc. This won’t be a valid assumption for the real case when we made the same circuit with material components. The values of measured currents and voltages won’t differ drastically; but yeah there will be some discrepancies with this simulation measurements.
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